Chapter 3: Problem 8
In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(f^{\prime}(g(x)) g^{\prime}(x)\) $$y=-\sec u, \quad u=\frac{1}{x}+7 x$$
Short Answer
Expert verified
\( \frac{dy}{dx} = (\sec(u) \tan(u)) (\frac{1}{x^2} - 7) \). Substitute \(u = \frac{1}{x} + 7x\) if necessary.
Step by step solution
01
Identify the given functions
We are given two functions: \(y = -\sec(u)\) and \(u = \frac{1}{x} + 7x\). Our task is to find \( \frac{dy}{dx} \).
02
Differentiate y with respect to u
To find \( \frac{dy}{dx} \), we first need \( \frac{dy}{du} \). Since \(y = -\sec(u)\), use the derivative of \(\sec(u)\) with respect to \(u\): \( \frac{d}{du}[ \sec(u) ] = \sec(u) \tan(u) \). Thus, \( \frac{dy}{du} = -( \sec(u) \tan(u) )\).
03
Differentiate u with respect to x
Next, we find the derivative of \(u\) with respect to \(x\): \(u = \frac{1}{x} + 7x\). Differentiate term by term: \( \frac{d}{dx} \left[ \frac{1}{x} \right] = -\frac{1}{x^2} \) and \( \frac{d}{dx}[7x] = 7 \). Thus, \( \frac{du}{dx} = -\frac{1}{x^2} + 7 \).
04
Apply the Chain Rule
Using the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). Substitute the derivatives calculated: \( \frac{dy}{dx} = \left( -(\sec(u) \tan(u)) \right) \times \left( -\frac{1}{x^2} + 7 \right) \).
05
Simplify the Expression
Simplify \( \frac{dy}{dx} = (\sec(u) \tan(u)) \times (\frac{1}{x^2} - 7) \). Substitute \(u = \frac{1}{x} + 7x\) if further substitution is needed for specific evaluations.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a fundamental concept in calculus. It deals with the idea of how a function changes as its input changes. In simpler terms, it's about finding the rate of change or the slope of a curve at any point. This rate of change is often referred to as the derivative. To perform differentiation, we apply specific rules or formulas that relate to the function's form.
- For example, if we have a function like \(y = x^2\), the derivative \(\frac{dy}{dx}\) gives us \(2x\), which tells us how the output \(y\) changes as \(x\) changes.
- The derivative of a linear function is constant, while for non-linear functions, it changes depending on \(x\).
Composite Functions
Composite functions occur when we combine two or more functions into a single function. The result is a new function that relies on the transformations of the original functions. For example, if a function \(u = g(x)\) is used inside another function \(y = f(u)\), we get a composite function \(y = f(g(x))\).
- This scenario often appears in calculus problems where the inner function \(g(x)\) modifies the input before it reaches the outer function \(f(u)\).
- The input \(x\) is first transformed by \(g(x)\) before \(f\) takes over, which affects the output \(y\).
Secant Function
The secant function, denoted as \(\sec(u)\), is a trigonometric function. It is the reciprocal of the cosine function, meaning \(\sec(u) = \frac{1}{\cos(u)}\). Here are some key points about the secant function:
- It is undefined wherever \(\cos(u) = 0\), leading to vertical asymptotes in its graph.
- Like other trigonometric functions, it repeats its pattern in a periodic way.